Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

From Ernie Cohen comes a scintillating stumper of a survey:

You’re reviewing some of the survey data that was randomly collected from the residents of Riddler City. As you’ll recall, the city is quite large.

Ten randomly selected residents were asked how many people (including them) lived in their household. As it so happened, their answers were 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

It’s your job to use this (admittedly limited) data to estimate the average household size in Riddler City. Your co-worker suggests averaging the 10 numbers, which would give you an answer of about 5.5 people. But you’re not so sure.

Would your best estimate be exactly 5.5, less than 5.5 or greater than 5.5?

The solution to this Riddler Express can be found in the following column.

## Riddler Classic

Congratulations, you’ve made it to the finals of the Riddler Ski Federation’s winter championship! There’s just one opponent left to beat, and then the gold medal will be yours.

Each of you will complete *two* runs down the mountain, and the times of your runs will be added together. Whoever skis in the least overall time is the winner. Also, this being the *Riddler* Ski Federation, you have been presented detailed data on both you and your opponent. You are evenly matched, and both have the same normal probability distribution of finishing times for each run. And for both of you, your time on the first run is completely independent of your time on the second run.

For the first runs, your opponent goes first. Then, it’s your turn. As you cross the finish line, your coach excitedly signals to you that you were faster than your opponent. Without knowing either exact time, what’s the probability that you will still be ahead after the second run and earn your gold medal?

*Extra credit:* Over in the snowboarding championship, there are 30 finalists, including you (apparently, you’re a dual-sport threat!). Again, you are the last one to complete the first run, and your coach signals that you are in the lead. What is the probability that you’ll win gold in snowboarding?

The solution to this Riddler Classic can be found in the following column.

## Solution to last week’s Riddler Express

Congratulations to 👏 Daniel Baker 👏 of Salt Lake City, Utah, winner of last week’s Riddler Express.

Last week, you were a contestant on the hit new game show, “You Bet Your Fife.” On the show, a random real number was chosen between 0 and 100. Your job was to guess a value that was *less than* this randomly chosen number. Your reward for winning was a novelty fife that was valued precisely at your guess. For example, if the number was 75 and you guessed 5, you would have won a $5 fife, but if you guessed 60, you would have won a $60 fife. Meanwhile, a guess of 80 would have won you nothing.

What number should you have guessed to maximize the average value of your fifing winnings?

If you guessed a smaller number, you were more likely to win a fife. But if you guessed a higher number — and you happened to win — you got a *more valuable* fife. With these competing priorities, some solvers reasoned that the answer should be right in the middle, with a guess of 50. But how could we show this mathematically?

One way was to determine the average winnings as a function of the number you guessed, which we’ll call *x*. This average was the product of your winnings (which would be *x*) and your probability of winning, which was (100−*x*)/100. In other words, this probability started at 100 percent when you guessed a value of zero, and linearly decreased to zero percent when you guessed a value of 100.

Putting these expressions together, a guess of *x* returned an average value of *x*(100−*x*)/100. To maximize your average winnings, you had to find the maximum of this quadratic function, which you could do with calculus (setting its derivative equal to zero), graphing or reasoning with symmetry. Any which way you did it, this maximum occurred when *x* was **50**.

Plugging in a value of 50 for *x* in the expression *x*(100−*x*)/100 meant that your average winnings were $25. Not too shabby. But if you didn’t win anything on “You Bet Your Fife,” you can always bid on a $100,000 flute.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Reuven 👏 of Ottawa, Ontario, Canada, winner of last week’s Riddler Classic.

Last week, the mysterious Barbara Yew offered a mysterious number puzzle, where you had to find eight three-digit numbers. The products of the three digits of each number were shown in the rightmost column of the table below. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, were shown in the bottom row.

First off, I’m delighted to report that this riddle was part of the annual (and this year, virtual!) MIT Mystery Hunt, puzzles from which have previously appeared in this column. The puzzle’s submitter, “Barbara Yew,” was a fictional character within the hunt, which was created and organized by last year’s winning team, the ✈✈✈ Galactic Trendsetters ✈✈✈. Indeed, this week’s winner, Reuven, participated in the hunt as part of team Control Group.

Your first step was to list out the different ways to multiply three digits from 1 to 9 to achieve each row’s product:

- 294 was 6×7×7.
- 216 was 3×8×9, 4×6×9 or 6×6×6.
- 135 was 3×5×9.
- 98 was 2×7×7.
- 112 was 2×7×8 or 4×4×7.
- 84 was 2×6×7 or 3×4×7.
- 245 was 5×7×7.
- 40 was 1×5×8 or 2×4×5.

For a few solvers, that last row — 40 — was tricky. You had to realize that 1 was an allowable digit, even though it isn’t a prime number.

From here, there were a total of 2,729,376 unique ways to place these digits in the table. That wasn’t a prohibitively large number, and so some solvers, like Siddhartha Srivastava of Patna, India, powered forth with computational brute force to find the unique solution. But this puzzle was indeed solvable with pen (okay, pencil), paper and logic.

Your next step was to find the prime factorization of the products of each column:

- 8,890,560 was 2
^{6}×3^{4}×5^{1}×7^{3}. - 156,800 was 2
^{7}×5^{2}×7^{2}. - 55,566 was 2
^{1}×3^{4}×7^{3}.

Solver Sara McArdle took a closer look at the first column, whose prime factorization implied that four of the eight digits had to be 5, 7, 7 and 7. That meant the remaining factorization, 2^{6}×3^{4}, or 5,184, had to be the product of the remaining four digits. There was only one way to break down 5,184 in this way: 8, 8, 9 and 9. The two factors of 9 had to come from the second (216) and third (135) rows. Also, because the third column did not have a factor of 5, that meant the three-digit number in the third row had to be 953.

Because all of the first column’s factors of 3 were accounted for, that meant the first row’s 6 couldn’t be in the first column. Also, the second column did not include a factor of 3, which meant the last column had the 6. In other words, the three-digit number in the first row was 776.

That 6 from the first row was responsible for the third column’s single factor of 2. Since the first digit of the second row was 9, the only way that row’s third digit could be odd was for the three-digit number in the second row to be 983.

But what about those two 8s in the first column? At this point, they had to be in the fifth and eighth rows. Because the last column had no factors of 5, the three-digit number in the eighth row had to be 851. And because the third column wasn’t allowed any more factors of 2, the three-digit number in the fifth row had to be 827.

The first column still needs a factor of 5, and the only remaining place it can get it is from the seventh row, whose three-digit number had to be 577. Meanwhile, the first column had all its factors of 2 accounted for, so the three-digit number of the fourth row had to be 727. Last but not least, the three-digit number of the sixth row was 743.

When all was said and done, here was the completed table:

Finally, a big congratulations to the winning team of this year’s incredible MIT Mystery Hunt, Palindrome!

## Want more riddles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com